4n^2-19n+18=

Solution for 4n^2-19n+18= equation:

4n^2-19n+18=

We move all terms to the left:

4n^2-19n+18-()=0

We add all the numbers together, and all the variables

4n^2-19n=0

a = 4; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·4·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*4}=\frac{0}{8}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*4}=\frac{38}{8}
=4+3/4
$